I've been trying to find the value of the following integral using the residue theorem : $$\int_0^\infty \frac{\sin ax\ dx}{(x^2+b^2)^2}$$
So basically the idea that I had is to transform this integral and to make it in the numerator as $e^{iaz} = \cos az\ + i\sin az\ $ and thus $$\mathrm{Im}\, \int_0^\infty \frac{e^{iaz}dz}{(z^2+b^2)^2}, = \int_0^\infty \frac{\sin az\ dz}{(z^2+b^2)^2}$$
However in class we've only established the residue formula from $-\infty$ to $+\infty$ , and when I try to integrate it that way (as in from $-\infty$ to $+\infty$) I obtain $0$ obviously because the function is odd.
- I'd like to add that I worked on a similair integral $\int_0^\infty \frac{\cos ax\ dx}{(x^2+b^2)^2}$ , using the poles $+ib$ and $-ib$, hence $(x^2+b^2)^2$ $=$ $(x+ib)^2(x-ib)^2$ and then calculating the $Res( \frac{\cos az\ dz}{(z^2+b^2)^2}, +ib)$ by using this : $$\mathrm{Re}\, \int_0^\infty \frac{e^{iaz}dz}{(z^2+b^2)^2}, = \int_0^\infty \frac{\cos az\ dz}{(z^2+b^2)^2}$$
I'm trying to do a similar thing with the sin function. calculating the cos integral was way easier due to the fact that the integrated is even.
Any help is welcome to direct me in the right path. Also apologies since english isn't my first language!